Leetcode 1376. Time Needed to Inform All Employees (python+cpp)

题目

解法1：BFS

class Solution:
    def numOfMinutes(self, n: int, headID: int, manager: List[int], informTime: List[int]) -> int:
        graph = collections.defaultdict(list)
        # s:subordinates,m:direct manager
        for s,m in enumerate(manager):
            graph[m].append(s)
        q = collections.deque()
        q.append((headID,0))
        ans = 0;
        
        while q:
            curr_m,time = q.popleft()
            if curr_m not in graph:
                ans = max(ans,time)
            for curr_s in graph[curr_m]:
                q.append((curr_s,informTime[curr_m]+time))
        
        return ans

C++版本
这边表示图既可以用unordered_map也可以用vector和array结合。unordered_map会稍微慢一点，因为他的查找操作并不是严格O(1)的

class Solution {
     
public:
    int numOfMinutes(int n, int headID, vector<int>& manager, vector<int>& informTime) {
     
        // constuct an array with size n, every element is a vector with variable size
        vector<int> graph[n];
        // unordered_map> graph; use this to construct the graph also works
        for(int i=0;i<manager.size();i++){
     
            if(manager[i] != -1){
     
                graph[manager[i]].push_back(i);
            }
            
        }
        
        queue<pair<int,int>> q;
        q.push({
     headID,0});
        int ans = 0;
        
        while(!q.empty()){
     
            pair<int,int> curr = q.front();
            q.pop();
            // if(graph.find(curr.first) == graph.end()) for map approach
            if(graph[curr.first].size() == 0){
     
                ans = max(ans,curr.second);
            }
            for(auto curr_s:graph[curr.first]){
     
                q.push({
     curr_s,curr.second+informTime[curr.first]});
            }
        }
        return ans;
    }
};

解法2：dfs

这道题dfs和bfs复杂度一致，每个节点都只需访问一遍

class Solution:
    def numOfMinutes(self, n: int, headID: int, manager: List[int], informTime: List[int]) -> int:
        def dfs(m,curr_time,max_time):
            if not graph[m]:
                return max(curr_time,max_time)
            for s in graph[m]:
                max_time= dfs(s,curr_time+informTime[m],max_time)
            return max_time
        graph = collections.defaultdict(list)
        for s,m in enumerate(manager):
            graph[m].append(s)
        
        return dfs(headID,0,0)